Question

find the equation of p, if the line segment joining (2,3) & (-1,5) substands a right angle at p.​

Answer 1

Given:-

• Coordinates of Q = (2,3)
• Coordinates of R = (-1,5)
• Angle QPR = 90°

To find:-

• the coordinates of P.

Solution :-

Let the coordinates of P be (x,y).

∠QPR = 90°

By using Pythagoras theorem in ∆PQR,

⇒ QR² = PQ² + PR²

⇒ (distance between QR)² = (distance between PQ)² + (distance between PR)²

⇒ (x – 2)² + (y – 3)² + (x + 1)² + (y – 5)² = (–1 – 2)² + (5 – 3)²

⇒x² +4 - 4x + y² + 9 - 6y + x² + 1 +2x + y² + 25 -10y = 9 + 4

⇒ 2x² + 2y² -16y -2x + 26 = 0

⇒ x² + y² – x – 8y + 13 = 0

∴ the locus of point P, having coordinates (x,y) is given by x² + y² – x – 8y + 13 = 0.

Related info:-

➸ Pythagoras theorem:

(Hypotenous)² = (perpendicular)² + (base)²

➸ Distance formula :

= (x2 - x1)² + (y2 - y1)²

➸ (a+b)² = a² + b² + 2ab

ㅤ(a-b)² = a² + b² - 2ab

Answer 2

Answer:

⭐ Question ⭐

✏ Find the equation of p, if the line segment joining (2,3) & (-1,5) substands a right angle at p.

⭐ Given ⭐

⚫ Coordinates of Q = (2,3)

⚫ Coordinates of R = (-1,5)

⚫ Angle QPR = 90°

⭐ To Find ⭐

✏ The coordinates of P.

⭐ Solution ⭐

✏ Let the coordinates of P be (x,y). ∠QPR = 90°

✏ By using Pythagoras theorem in ∆PQR

=> QR² = PQ² + PR²

=> (x – 2)² + (y – 3)² + (x + 1)² +

(y – 5)² = (-1– 2)²+ (5 – 3)²

=> x²+4 - 4x + y² + 9 - 6y + x² + 1 +2x + y²+ 25 -10y = 9 + 4

=> 2x² + 2y² -16y -2x + 26 = 0

=> x²+ y²– x - 8y + 13 = 0

=> The locus of point P, having coordinates (x,y) is given by x² + y²- x- 8y + 13 = 0.

Step-by-step explanation:

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